Question

O is any point in interior of rectangle ABCD prove that

Answer 1

$\large\bf{\underline{\underline{Correct\:Question:−}}}$

A point O is in the interior of a rectangle ABCD, is joined with each of the vertices A, B, c & D Prove that OA2 +OC2 08+00

$\large\bf{\underline{\underline{Answer :-}}}$

$\underline{\boxed{\sf\purple{We \:draw \:PQ \:║\:AB\: ║\:CD}}}$

ABCD is a rectangle , it means ABPQ and PQDC are also rectangle

$\sf{For\: , ABPQ ,}$

$\underline{\boxed{\sf\purple{}}}$AP = BQ [opposite sides are equal ]

$\sf{For,\: PQDC}$

$\underline{\boxed{\sf\purple{PD \:=\: QC}}}$ $\tiny\sf\color{red}{[ opposite\: sides\: are \:equal ]}$

$\sf{now, \:for \:∆OPD,}$

$\underline{\boxed{\sf\purple{OD² = OP² + PD² ------(1)}}}$

$\sf{For,\: ∆OQB,}$

$\underline{\boxed{\sf\purple{OB² = OQ² + BQ² --------(2)}}}$

$\textbf{Add equations (1) and (2),}$

$\color{blue}\boxed{\sf{OB² + OD² = (OP² + PD²)+ (OQ² + BQ²)}}$

$\color{blue}\boxed{\sf{= (OP² + CQ²) + (OQ² + AP²)}}$

$\small\textbf{∆OPA and ∆OQC are also right angled triangles}$

$\underline{\boxed{\sf\purple{For, \:∆OCQ}}}$

$\rm{⇒OQ² + CQ² = OC²}$

$\underline{\boxed{\sf\purple{For,\:∆OPA}}}$

$\rm{⇒OP² + AP² = OA²}$

$\underline{\boxed{\sf\purple{Now ,}}}$

$\red{\bf {OB² \:+\: OD² \:= \:OC² \:+\: OA²}}$

$\sf{Hence,\: proved}$

Thankyou :)